Question

Energy can be removed from water as heat at and even below the normal freezing point ($\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$at atmospheric pressure) without causing the water to freeze; the water is then said to be supercooled. Suppose a 1.00 gwater drop is super-cooled until its temperature is that of the surrounding air, which is at$\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{°}\mathbf{C}$. The drop then suddenly and irreversibly freezes, transferring energy to the air as heat. What is the entropy change for the drop? (Hint: Use a three-step reversible process as if the water were taken through the normal freezing point.) The specific heat of ice is$\mathbf{2220}\mathbf{}\mathit{J}\mathbf{/}\mathit{k}\mathit{g}\mathbf{.}\mathit{K}$.

Short answer

Entropy change for the drop is$-1.18\mathrm{J}/\mathrm{K}$

Step-by-step solution

The given data

Mass of water drop is$\mathrm{m}=1\mathrm{gm}$

The initial temperature of water drop${\mathrm{T}}_1=-5°\mathrm{C}=268\mathrm{K}$

Temperature${\mathrm{T}}_2=0°\mathrm{C}=273\mathrm{K}$

Temperature${\mathrm{T}}_3={\mathrm{T}}_2$

Temperature${\mathrm{T}}_4={\mathrm{T}}_1$

Use a three-step reversible process as if the water were through the normal freezing point.

Understanding the concept of entropy change

We use the three-step reversible process to calculate the net entropy change of the water drop.

Formula:

The entropy change of gas for a reversible process,

$∆\mathrm{S}=\mathrm{mCIn}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$ (1)

Calculation of the change in entropy

We consider the three-step reversible process: The super-cooled water drop (of mass) starts at state1 $\left({\mathrm{T}}_1=268\mathrm{K}\right)$moves on to state 2 (still in liquid form but at temperature ${\mathrm{T}}_2=273\mathrm{K}$), freezes to state 3 $\left({\mathrm{T}}_3={\mathrm{T}}_1\right)$, and cools down to state 4 (in solid form with temperature ${\mathrm{T}}_4={\mathrm{T}}_1$). The change in entropy for each stage using equation (1) is given as below:

$∆\mathrm{S}={\mathrm{mC}}_{\mathrm{w}}\mathrm{In}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)$

$$\begin{gathered} ∆{\mathrm{S}}_{23}=-\frac{{\mathrm{mL}}_{\mathrm{F}}}{{\mathrm{T}}_2} \\ ={\mathrm{mC}}_{\mathrm{l}}\mathrm{In}\left(\frac{{\mathrm{T}}_4}{{\mathrm{T}}_2}\right) \\ ={\mathrm{mC}}_{\mathrm{l}}\mathrm{In}\left(\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\right) \\ ={\mathrm{mC}}_{\mathrm{l}}\mathrm{In}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right) \end{gathered}$$

Thus, the net entropy change for the drop using the given values is given as follows:

$$\begin{gathered} ∆\mathrm{S}=∆{\mathrm{S}}_{12}+∆{\mathrm{S}}_{23}+∆{\mathrm{S}}_{34} \\ ={\mathrm{mC}}_{\mathrm{w}}\mathrm{In}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)-\frac{{\mathrm{mL}}_{\mathrm{F}}}{{\mathrm{T}}_2} \\ =\mathrm{m}\left({\mathrm{C}}_{\mathrm{w}}-{\mathrm{C}}_{\mathrm{i}}\right)\mathrm{In}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)-\frac{{\mathrm{mL}}_{\mathrm{F}}}{{\mathrm{T}}_2} \\ =1\mathrm{gm}\left(4.19\mathrm{J}/\mathrm{gm}··\mathrm{K}-2.22\mathrm{J}/\mathrm{gm}·\mathrm{K}\right)\mathrm{In}\left(\frac{273\mathrm{K}}{268\mathrm{K}}\right)-\frac{\left(1\mathrm{gm}\right)\left(333\mathrm{J}/\mathrm{gm}\right)}{273\mathrm{K}} \\ =\left[\left(1.97\right)\left(0.01848\right)-1.2197\right]\mathrm{J}/\mathrm{K} \\ =-1.18\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of the gas is$-1.18\mathrm{J}/\mathrm{K}$